A123760 -id:A123760 - cp's OEIS frontend

A093873 Numerators in Kepler's tree of harmonic fractions.

1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 3, 1, 3, 2, 3, 1, 4, 3, 4, 2, 5, 3, 5, 1, 4, 3, 4, 2, 5, 3, 5, 1, 5, 4, 5, 3, 7, 4, 7, 2, 7, 5, 7, 3, 8, 5, 8, 1, 5, 4, 5, 3, 7, 4, 7, 2, 7, 5, 7, 3, 8, 5, 8, 1, 6, 5, 6, 4, 9, 5, 9, 3, 10, 7, 10, 4, 11, 7, 11, 2, 9, 7, 9, 5, 12, 7, 12, 3, 11, 8, 11, 5, 13, 8, 13, 1, 6

Offset: 1

N. J. A. Sloane and Reinhard Zumkeller, May 24 2004

Form a tree of fractions by beginning with 1/1 and then giving every node i/j two descendants labeled i/(i+j) and j/(i+j).

			The first few fractions are:
1 1 1 1 2 1 2 1 3 2 3 1 3 2 3 1 4 3 4 2 5 3 5 1 4 3 4 2 5 3 5
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ...
1 2 2 3 3 3 3 4 4 5 5 4 4 5 5 5 5 7 7 7 7 8 8 5 5 7 7 7 7 8 8

R. Zumkeller, Table of n, a(n) for n = 1..10000
Johannes Kepler, Harmonices Mundi, Liber III, see p. 27.
Index entries for fraction trees
Index entries for sequences related to music

The denominators are in A093875. Usually one only considers the left-hand half of the tree, which gives the fractions A020651/A086592. See A086592 for more information, references to Kepler, etc.

See A294442 for another version of Kepler's tree of fractions.

{-# LANGUAGE ViewPatterns #-}
import Data.Ratio((%), numerator, denominator)
rat :: Rational -> (Integer,Integer)
rat r = (numerator r, denominator r)
data Harmony = Harmony Harmony Rational Harmony
rows :: Harmony -> [[Rational]]
rows (Harmony hL r hR) = [r] : zipWith (++) (rows hL) (rows hR)
kepler :: Rational -> Harmony
kepler r = Harmony (kepler (i%(i+j))) r (kepler (j%(i+j)))
           where (rat -> (i,j)) = r
-- Full tree of Kepler's harmonic fractions:
k = rows $ kepler 1 :: [[Rational]] -- as list of lists
h = concat k :: [Rational] -- flattened
a093873 n = numerator $ h !! (n - 1)
a093875 n = denominator $ h !! (n - 1)
a011782 n = numerator $ (map sum k) !! n -- denominator == 1
-- length (k !! n) == 2^n
-- numerator $ (map last k) !! n == fibonacci (n + 1)
-- denominator $ (map last k) !! n == fibonacci (n + 2)
-- numerator $ (map maximum k) !! n == n
-- denominator $ (map maximum k) !! n == n + 1
-- eop.
-- Reinhard Zumkeller, Oct 17 2010

M:= 8: # to get a(1) .. a(2^M-1)
gen[1]:= [1];
for n from 2 to M do
  gen[n]:= map(t -> (numer(t)/(numer(t)+denom(t)),
      denom(t)/(numer(t)+denom(t))), gen[n-1]);
od:
seq(op(map(numer,gen[i])),i=1..M): # Robert Israel, Jan 11 2016

num[1] = num[2] = 1; den[1] = 1; den[2] = 2; num[n_?EvenQ] := num[n] = num[n/2]; den[n_?EvenQ] := den[n] = num[n/2] + den[n/2]; num[n_?OddQ] := num[n] = den[(n-1)/2]; den[n_?OddQ] := den[n] = num[(n-1)/2] + den[(n-1)/2]; A093873 = Table[num[n], {n, 1, 97}] (* Jean-François Alcover, Dec 16 2011 *)

a(n) = a([n/2])*(1 - n mod 2) + A093875([n/2])*(n mod 2).
a(A029744(n-1)) = 1; a(A070875(n-1)) = 2; a(A123760(n-1)) = 3. - Reinhard Zumkeller, Oct 13 2006
A011782(k) = SUM(a(n)/A093875(n): 2^k<=n<2^(k+1)), k>=0. [Reinhard Zumkeller, Oct 17 2010]
a(1) =  1. For all n>0  a(2n) =  a(n), a(2n+1) =  A093875(n). - Yosu Yurramendi, Jan 09 2016
a(4n+3) = a(4n+1), a(4n+2) = a(4n+1) - a(4n), a(4n+1) = A071585(n). - Yosu Yurramendi, Jan 11 2016
G.f. G(x) satisfies G(x) = x + (1+x) G(x^2) + Sum_{k>=2} x (1+x^(2^(k-1))) G(x^(2^k)). - Robert Israel, Jan 11 2016
a(2^(m+1)+k) = a(2^(m+1)+2^m+k) = A020651(2^m+k), m>=0, 0<=k<2^m. - Yosu Yurramendi, May 18 2016
a(k) = A020651(2^(m+1)+k) - A020651(2^m+k), m>0, 0Yosu Yurramendi, Jun 01 2016
a(2^(m+1)+k) - a(2^m+k) = a(k) , m >=0, 0 <= k < 2^m. For k=0 a(0)=0 is needed. - Yosu Yurramendi, Jul 22 2016
a(2^(m+2)-1-k) = a(2^(m+1)-1-k) + a(2^m-1-k), m >= 1, 0 <= k < 2^m. - Yosu Yurramendi, Jul 22 2016
a(2^m-1-(2^r -1)) = A000045(m-r), m >= 1, 0 <= r <= m-1. - Yosu Yurramendi, Jul 22 2016
a(2^m+2^r) = m-r, , m >= 1, 0 <= r <= m-1 ; a(2^m+2^r+2^(r-1)) = m-(r-1), m >= 2, 0 <= r <= m-1. - Yosu Yurramendi, Jul 22 2016
A093875(2n) - a(2n) = A093875(n), n > 0; A093875(2n+1) - a(2n+1) = a(n), n > 0. - Yosu Yurramendi, Jul 23 2016

A070875 Binary expansion is 1x100...0 where x = 0 or 1.

Original entry on oeis.org

5, 7, 10, 14, 20, 28, 40, 56, 80, 112, 160, 224, 320, 448, 640, 896, 1280, 1792, 2560, 3584, 5120, 7168, 10240, 14336, 20480, 28672, 40960, 57344, 81920, 114688, 163840, 229376, 327680, 458752, 655360, 917504, 1310720, 1835008, 2621440

Offset: 0

N. J. A. Sloane, May 19 2002

A 2-automatic sequence. - Charles R Greathouse IV, Sep 24 2012
Third row in array A228405. - Richard R. Forberg, Sep 06 2013
Conjecture: a(n) = -1 +  positions of the ones in A309019(n+2) - A002487(n+2). - George Beck, Mar 26 2022
Consecutive integers for which the number of its proper nondivisors of the form 2^k (k > 0) is 2; proper nondivisors are defined in A173540 (5 has two such nondivisors: 2 and 4, 7 has 2 and 4, 10 has 4 and 8, 14 has 4 and 8, 20 has 8 and 16,...). - Lechoslaw Ratajczak, Dec 17 2024

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2).
Index entries for 2-automatic sequences.

Cf. A070876, A123760, A063920.

[n le 2 select 2*n+3 else 2*Self(n-2): n in [1..39]]; // Bruno Berselli, Mar 01 2011

Flatten@ NestList[ 2# &, {5, 7}, 19] (* Or *)
CoefficientList[ Series[(5 + 7 x)/(1 - 2 x^2), {x, 0, 38}], x] (* Robert G. Wilson v, May 20 2002 *)

a(n)=if(n%2,7,5)<<(n\2) \\ Charles R Greathouse IV, Sep 24 2012

A093873(a(n)) = 2. - Reinhard Zumkeller, Oct 13 2006
For n>1, a(n+1) = a(n) + A000010(a(n)). - Stefan Steinerberger, Dec 20 2007
From Bruno Berselli, Mar 01 2011: (Start)
G.f.: (5+7*x)/(1-2*x^2).
a(n) = (6-(-1)^n)*2^((2*n+(-1)^n-1)/4). Therefore: a(n) = 5*2^(n/2) for n even, otherwise a(n) = 7*2^((n-1)/2).
a(n) = 2*a(n-2) for n>1. (End)
a(n+1) = A063757(n) + 6. - Philippe Deléham, Apr 13 2013
a(n) = sqrt(2*a(n-1) - (-2)^(n-1)). - Richard R. Forberg, Sep 06 2013
a(n+3) = a(n+2)*a(n+1)/a(n). - Richard R. Forberg, Sep 06 2013
For n>1, a(n) = 2*phi(a(n)) + phi(phi(a(n))). - Ivan Neretin, Feb 28 2016
a(2n) = A020714(n), a(2n+1) = A005009(n); for n>0. - Yosu Yurramendi, Jun 01 2016
From Ilya Gutkovskiy, Jun 02 2016: (Start)
E.g.f.: 7*sinh(sqrt(2)*x)/sqrt(2) + 5*cosh(sqrt(2)*x).
a(n) = 2^((n-3)/2)*(5*sqrt(2)*(1 + (-1)^n) + 7*(1 - (-1)^n)). (End)
Sum_{n>=0} 1/a(n) = 24/35. - Amiram Eldar, Mar 28 2022

Extended by Robert G. Wilson v, May 20 2002

A265027 First differences of A048701 divided by 6.

Original entry on oeis.org

1, 1, 3, 2, 1, 2, 11, 4, 2, 4, 1, 4, 2, 4, 43, 8, 4, 8, 2, 8, 4, 8, 1, 8, 4, 8, 2, 8, 4, 8, 171, 16, 8, 16, 4, 16, 8, 16, 2, 16, 8, 16, 4, 16, 8, 16, 1, 16, 8, 16, 4, 16, 8, 16, 2, 16, 8, 16, 4, 16, 8, 16, 683, 32, 16, 32, 8, 32, 16, 32, 4, 32, 16, 32, 8, 32, 16, 32, 2, 32, 16, 32, 8, 32, 16, 32, 4, 32, 16, 32, 8

Offset: 2

N. J. A. Sloane, Nov 30 2015

nonn

Indices n such that a(n) = 1 are equal to row sums of Lucas triangle. In other words, a(A042950(n)) = 1. Additionally, a(A070875(n)) = 2 and a(A123760(n)) = 4. - Altug Alkan, Dec 04 2015

N. J. A. Sloane, Table of n, a(n) for n = 2..32767

Cf. A042950, A048701, A070875, A123760, A265026.

Differences@ Select[Range@ 12000, Reverse@ # == # && EvenQ@ Length@ # &@ IntegerDigits[#, 2] &]/6 (* Michael De Vlieger, Dec 04 2015 *)

a048701(n) = my(f); f = length(binary(n)) - 1; 2^(f+1)*n + sum(i=0, f, bittest(n, i) * 2^(f-i));
vector(100, n, (a048701(n+1) - a048701(n)) / 6) \\ Altug Alkan, Dec 03 2015

a(n) = A265026(n) / 6, for n > 1. - Altug Alkan, Dec 03 2015

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A093873 Numerators in Kepler's tree of harmonic fractions.

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A070875 Binary expansion is 1x100...0 where x = 0 or 1.

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A265027 First differences of A048701 divided by 6.

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